Testing Equivalence of Polynomials under Shifts

Abstract

Two polynomials f, g ∈ F[x 1 , . . . , x n ] are called shift-equivalent if there exists a vector (a 1 , . . . , a n ) ∈ F n such that the polynomial identity f (x 1 + a 1 , . . . , x n + a n ) ≡ g(x 1 , . . . , x n ) holds. Our main result is a new randomized algorithm that tests whether two given polynomials are shift equivalent. Our algorithm runs in time polynomial in the circuit size of the polynomials, to which it is given black box access. This complements a previous work of Grigoriev [Gri97] who gave a deterministic algorithm running in time n O(d) for degree d polynomials.Our algorithm uses randomness only to solve instances of the Polynomial Identity Testing (PIT) problem. Hence, if one could de-randomize PIT (a long-standing open problem in complexity) a de-randomization of our algorithm would follow. This establishes an equivalence between de-randomizing shift-equivalence testing and de-randomizing PIT (both in the black-box and the white-box setting). For certain restricted models, such as Read Once Branching Programs, we already obtain a deterministic algorithm using existing PIT results.

Existence Problem of Telescopers for Rational Functions in Three Variables

“…This problem is solved by Grigoriev in [20,21] and more recently by Dvir et al in [18] with better complexity. Problem 4.4 (Separation Problem).…”

Reconstruction of Full Rank Algebraic Branching Programs

“…Two n variate degree d polynomials f, g ∈ F[x] are translation equivalent (also called shift equivalent in [13]) if there exists a point a ∈ F n such that f (x + a) = g(x). Translation equivalence test takes input blackbox access to two n variate polynomials f and g, and outputs an a ∈ F n such that f (x + a) = g(x) if f and g are translation equivalent else outputs 'f and g are not translation equivalent'.…”

“…Translation equivalence test takes input blackbox access to two n variate polynomials f and g, and outputs an a ∈ F n such that f (x + a) = g(x) if f and g are translation equivalent else outputs 'f and g are not translation equivalent'. As before, let β be the bit lengths of the coefficients of f and g. A randomized poly(n, d, β) time algorithm is presented in [13] to test translation equivalence and find an a ∈ F n such that f (x + a) = g(x), if such an a exists. Another randomized test was mentioned in [25], which we present as proof of the following lemma in Section 7.1.…”

Existence Problem of Telescopers: Beyond the Bivariate Case

“…x i on K[x 1 , ..., x n ]. Given p ∈ K[x 1 , ..., x n ], to decide whether there exist integers m 1 , ..., m n with m 1 > 0 such that σ m1 1 • • • σ mn n (p) = p. This problem is a special case of the problem proposed and solved by Grigoriev in [19,20] and more recently by Dvir et al in [17]. Theorems 6.4 and 6.6 reduce the problem to that of testing the summability of bivariate rational functions.…”